# What does $e^{\mu}$ mean for a measure $\mu$?

I have seen the notation $\int_M fe^{\mu}$ in some geometry books and I cannot even guess what $e^{\mu}$ might mean for a measure/form $\mu$ on the (symplectic) manifold $M$.

Any clarifications are appreciated.

Edit

• Where did you see this notation?
– user99914
Commented Nov 10, 2014 at 5:42
• @John I included an example from repn theory/symp geom. Commented Nov 10, 2014 at 6:27
• Commented Nov 10, 2014 at 6:44
• @JairTaylor what about Esigma? Commented Nov 10, 2014 at 6:50
• My only thought is $e^\sigma = e^{(\sigma/\omega)} \omega$ for $\omega$ the canonical volume form, but I have never seen the notation before, nor studied symplectic geometry. Do none of the books define it? Commented Nov 10, 2014 at 7:12

I found this remark in a book by Guillemin, Ginzburg, and Karshon:

It is convenient to work with the differential form (of mixed degree) $$\exp \omega=1+\omega+\frac{1}{2!}\omega\wedge\omega\dots .$$ With the convention that $\int_M\beta=0$ if the degree of $\beta$ is different than the dimension of $M$, Liouville measure is given by integration of $\exp \omega$.

This shows that my guess in the above comments was correct! However, I would appreciate if an expert would clarify why it is "convenient" to work with this differential form of mixed degrees.

• I guess because it's easier to write $e^\omega$ than $\frac{1}{7!} \omega\wedge\omega \wedge \dotsc \wedge\omega$. But that's just a guess. Commented Nov 10, 2014 at 22:21
• @DanielFischer I don't think writing $e^{\omega}$ is so much easier than $\omega^n/n!$ that justifies its use. Commented Nov 10, 2014 at 22:59
• This remind me of Chern character en.wikipedia.org/wiki/Chern_class
– user99914
Commented Nov 11, 2014 at 2:58

If $e^{\mu}=d\lambda$ for some measure $\lambda$ in $M$ then, it seems that $\mu=\ln(d\lambda)$. But here is not clear whether $\ln(d\lambda)$ is a measure.